Chemistry 9th Edition

Published by Cengage Learning
ISBN 10: 1133611095
ISBN 13: 978-1-13361-109-7

Chapter 5 - Gases - Exercises: 71

Answer

$33.2\text{ grams $KrCl_4$}$

Work Step by Step

Reaction: $$Kr+2Cl_2\Rightarrow KrCl_4$$ Find the limiting reactant: $PV=nRT$ $0.5*15.0=n(0.08206)(623)$ $n=0.147\text{ mol Kr}$ $1.5*15.0=n*0.08206*623$ $n=0.4401\text{ mol $Cl_2$}$ Kr is the limiting reactant. $0.147\text{ mol Kr}*\frac{1\text{ mol $KrCl_4$}}{1\text{ mol Kr}}*\frac{225.6100\text{ grams $KrCl_4$}}{1\text{ mol $KrCl_4$}}=33.2\text{ grams $KrCl_4$}$
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